Are all values of $x$ solutions for $e^{2\ln(\sin(x))} = 1 - e^{2\ln(\cos(x))}$ in $\mathbb R$?

Question

Does all values of $x$ in $\mathbb R$ satisfy equation: $$e^{2\ln(\sin(x))} = 1 - e^{2\ln(\cos(x))}$$

I am asking this, because by checking WolframAlpha solution there is an answer: (all values for $x$ are solutions over reals), but we know that $\ln(0)$ is undefined, same for negative numbers.

Wolfram Alpha solution

Therefore I assume in $\mathbb R$, zero and negative numbers doesn't satisfy this equation.

Answer 1

$$1=e^{2\ln\sin(x)} +e^{2\ln \cos(x)}= e^{\ln\sin^2(x)}+e^{\ln \cos^2(x)} = $$

$$ = \sin^2(x)+\cos^2(x) = 1 $$

So this is true for all $x$ such that $\sin x>0$ and $\cos x>0$ (because then is $\ln $ defined).

So $$\boxed{x\in \color{red}{\bigcup_{k\in \mathbb{Z}} (0+2\pi k,{\pi\over 2}+2\pi k)}}$$

Answer 2

Note that : $$e^{2\ln \sin x} + e^{2 \ln \cos x} = 1 \Rightarrow e^{\ln (\sin x)^2} + e^{\ln (\cos x)^2} = 1 \Leftrightarrow \sin^2x + \cos^2x = 1 \rightarrow \text{true} \; \forall x \in \mathbb R$$ Restrictions apply so as the initial expression holds, so that narrows down the solution set. Note the usage of $\Rightarrow$ instead of an $\Leftrightarrow$ at the start.